Answer
For: $$f(x)=\frac{x+2}{x}$$
$Domain :$$$ Domain: (0,\infty) and (-\infty,0)$$
f(x) has defined values for all x except x=0 $$
$$ Range:$$ (\infty,1) ,and, (-\infty,1)$$
For:
$$f(x)^{-1}= \frac{2}{x-1}$$
$$ Domain: x\ne1 $$ all real numbers except x≠1
$$ Range: (1,\infty) ,and, (1,-\infty)$$
Work Step by Step
Step1: Given$$f(x)=\frac{x+2}{x}$$
Step2: Simplify
$$f(x)=\frac{x+2}{x}=1+\frac{2}{x}$$
Step3: Defining Domain and Range
$$f(x)=\frac{x+2}{x}=1+\frac{2}{x}$$
$$ Domain: (0,\infty) and (-\infty,0)$$
f(x) has defines values for all x except x=0 $$
$$ Range:$$ (\infty,1) ,and, (-\infty,1)$$
$\lim\limits_{x \to -0}1+\frac{2}{x}
=-\infty$,and $\lim\limits_{x \to +0}1+\frac{2}{x}
=+\infty$
Likewise,
$\lim\limits_{x \to -\infty}1+\frac{2}{x}
=1$,and $\lim\limits_{x \to +\infty}1+\frac{2}{x}
=1$
when x is very small, such as 0.0000001,and, -0.0000001 f(x)= large and undefined, but, when x is large, such as 1,000,0000,and -1, 000,0000 then f(x)= 1.
Step4: Finding Inverse- Switching the places x and f(x)
$$x=1+\frac{2}{f(x)}$$
$$f(x)^{-1}= \frac{2}{x-1}$$
$ solving for f(x) $
Step5: Domain of $f(x)^{-1}$
$$ Domain: x\ne1 $$ all real numbers except x≠1
Step6: Range of $f(x)^{-1}$
$$\lim\limits_{x \to1^{-} }\frac{2}{x-1}=-\infty$$
$$\lim\limits_{x \to\infty^{-} }\frac{2}{x-1}=0$$
$$\lim\limits_{x \to1^{+} }\frac{2}{x-1}=+\infty$$
$$\lim\limits_{x \to\infty^{+} }\frac{2}{x-1}=0$$
